Growth vs Decay Math Example 3

Follow the full solution, then compare it with the other examples linked below.

Example 3

easy
A population grows by 8%8\% per year. Starting at 50005000, write the growth function and find the population after 1010 years.

Solution

  1. 1
    Growth rate r=0.08r=0.08, so base b=1+0.08=1.08b=1+0.08=1.08. Function: P(t)=5000โ‹…(1.08)tP(t)=5000\cdot(1.08)^t.
  2. 2
    P(10)=5000โ‹…(1.08)10โ‰ˆ5000โ‹…2.1589โ‰ˆ10,795P(10)=5000\cdot(1.08)^{10}\approx5000\cdot2.1589\approx10{,}795.

Answer

P(t)=5000โ‹…(1.08)tP(t)=5000\cdot(1.08)^t; P(10)โ‰ˆ10,795P(10)\approx10{,}795
An annual growth rate of r%r\% corresponds to multiplying by 1+r/1001+r/100 each year. After tt years, the factor (1+r/100)t(1+r/100)^t compounds the growth.

About Growth vs Decay

Exponential growth occurs when a quantity multiplies by a factor >1> 1 repeatedly; exponential decay when it multiplies by a factor between 0 and 1.

Learn more about Growth vs Decay โ†’

More Growth vs Decay Examples